Search All of the Math Forum:
Views expressed in these public forums are not endorsed by
NCTM or The Math Forum.


jaakov
Posts:
11
Registered:
12/3/12


Re: Given a set , is there a disjoint set with an arbitrary cardinality?
Posted:
Dec 3, 2012 11:41 AM


On 03.12.2012 17:33, Aatu Koskensilta wrote: > jaakov<removeit_jaakov@deleteit_ro.ru> writes: > >> Thank you. To finish this proof, we have to prove that your class, let >> us call it Z, contains an initial segment of cardinality k. However, >> informally speaking, X may hypothetically contain too many ordinals, >> still being a set, such that too few ordinals remain in Z. Could you >> please expand on why Z contains an initial segment of cardinality k? > > Assuming choice, the cardinality k of X is an (initial) ordinal, so > take the set X' = {alpha  alpha< k} and let Y be the set {x + lambda  > x in X'}, where lambda is an ordinal> any ordinal in X. > 1. k is not related to the cardinality of X.
2. Your lambda need not exist. To show that lambda exists, one has to show that a set may not contain arbitrarily large ordinals. Is it a known fact or simply false?
Jaakov.
 news://freenews.netfront.net/  complaints: news@netfront.net 



